Chapter 3 Steady-State, One-Dimensional i Conduction
3.1 The Plane Wall 3.1.1 Temperature Distribution For one-dimensional, steady-state conduction in a plane wall with no heat generation, the differential equation (.13) reduces to d ( dx k dt dx)= 0 (3.1) and the heat flux is a constant, independent of x. With further assumption of constant t, we have the general linear solution T(x) = C 1 x + C (3.) The heat flux is q qx k x = A = L (T s,1 T s, ) (3.5)
3.1. Thermal Resistance With the analogy between the diffusion of heat and electrical charge, the thermal resistance for conduction ction is R t,cond = T s,1 T s, q x = ka L (3.6) Similar il for convection with Newton's law of cooling R t,conv = T s T = 1 ha (3.9) s q In Fig. 3.1, the total thermal resistance, R tot,is R tot = h 1 + L + 1 1 A ka h A (3.1)
3.1.3 The Composite Wall (Fig. 3.; Fig. 3.3) 3) R tot = R t = ΔT q = 1 UA where U is the overall heat transfer coefficient, defined by analogy to Newton's law of cooling as q x UAΔT (3.19)
3.1.4 Contact Resistance --significant temperature drop exists across an interface due to the gaps between een the contact area Contact resistance can be reduced by increasing the joint pressure, reducing the roughness of the mating surfaces, applying a metal coating, inserting the interface with soft metal foil, or filling fluid of large thermal conductivity, it etc. See Tables 3.1 and 3..
EX 3.1, 3., 3.3
3.33 Radial Systems 3.3.1 The Cylinder For steady-state conditions with no heat generation, Eq..0 is reduced to 1 d ( ) r dr kr dt dr )= 0 (3.3) The conduction heat transfer rate q (not the heat flux q r r ) is a constant in the radial direction q r = ka dt dr = k(πrl) dt dr (3.4) The general solution of (3.3) is T(r) = C 1 ln r + C (3.5) An example is shown in Fig. 3.6.
T(r) = T T s,1 s, ln(r 1 / r ) ln r r + T s, (3.6) q r = πlk(t T ) s,1 s, (3.7) ln(r / r 1 ) ln(r / r ) 1 R t,cond = πlk (3.8) The temperature distribution for a composite cylindrical wall is shown in Fig. 3.7. EX 3.5
3.5 Conduction with Thermal Energy Generation 3.5.1 The Plane Wall For steady-state conditions with constant k and uniform energy generation per unit volume, Eq..16 becomes d T dx q& + k = 0 The general solution is q& T ( x) = x + C1x + C k (3.39) Some examples are shown in Fig. 3.9. With heat generation the heat flux is no longer independent of x. EX 3.7
3.6 Heat Transfer from Extended Surfaces Use of fins to enhance heat transfer from a wall: Figs. 3.1-3.14 3.6.1 A General Conduction Analysis (Fig. 3.15) Through energy balance, we obtain or d ( dx A dt c dx ) k h d T dx + 1 A c da c dx da s dx (T T ) = 0 dt dx 1 A c h k da s dx (T T ) = 0 (3.61)
θ(0) = T b T θ b 3.6. Fins of Uniform Cross-Sectional Area With da c /dx = 0 and da s /dx = P, Eq. 3.61 reduces to d T dx hp (T T ka ) = 0 (3.6) c Defining an excess temperature θ as θ(x) T (x) T (3.63) d θ m θ = 0 where m hp dx ka c (3.64,65) The general solution of (3.64) is θ(x) = C 1 e mx + C e mx (3.66) Two boundary conditions are needed. One boundary condition can be specified at the base of the fin (x = 0) as θ(0) ( ) = T b T θ b
The second condition may correspond to one of the four physical situations shown in Table 3.4. The solution procedure to obtain temperature distribution θ/θ b and fin heat transfer rate q f is discussed in the textbook. EX 3.9
Proper Length of a Fin Heat transfer ratio between a fin of finite length L and a fin of q f, infinite length = L (3.76) =tanhml q f, (3.80) fin length L <.65/m L In practice, fin length is usually constrained by space and weight. 3.6.3 Fin Performance ml Fin effectiveness ε tanhml f : ratio of the fin heat transfer 0.1 0.100 rate to that without the fin 0.5 0.46 q f 1.0 0.76 ε = f ha.0 0.964 c bθ (3.81), b 5.5 0.987 Use of fins may rarely be justified unless ε f γ. ε kp f hac 1/.65 0.990 3.0 0.995 ε (3.8) 5.0 1.000 Consider an infinite fin, assuming h unaltered by the presence of fin, ε f as k, P/A c (high k material; thin fins) h (gas medium and/or natural convection)
Fin resistance R tf θ b t,f : R t, f = qf (3.83) Fin efficiency η f : ratio of q f and the maximum rate when the entire fin surface were at the base temperature q f q f η f = qmax ha f θ b (3.86) For a straight fin of uniform cross section and an adiabatic tip, Eqs. 3.76 and 3.86 yield η M tanh ml tanh ml f = = hplθ b ml (3.87) This expression can be applied to a fin with convective tip, if a corrected tdfin length thl L c (eg., L c = L + t/) for a rectangular fin) is used. The fin efficiency is shown in Fig. 3.18. The fin resistance can be written in terms of η f. R t, = 1, f ha f η f (3.9) The η f and A f for some examples are shown in Table 3.5.
364FinsofNonuniform 3.6.4 Cross-Sectional Area For an annular fin with uniform fin thickness, d T dr + 1 r dt dr h kt (T T ) = 0 with m =h/kt h/kt, θ = T-T, d θ + 1 dθ m θ = 0 -- modified Bessel eq. of order zero dr r dr The general solution is of the form θ(r) ) = C 1 I 0 (mr) + C K 0 (mr) where I 0 and K 0 are modified, zero-order Bessel functions of the first and second kinds, respectively. <Home Work> Solve for the expression of θ/θ b, given the on p.151, for the case with a given base temperature and an adiabatic tip. The fin efficiencies of some examples with nonuniform cross- The fin efficiencies of some examples with nonuniform crosssectional area are shown in Figs. 3.18 and 3.19.
About Modified Bessel Equation Modified Bessel Equation of order v d θ v + 1 dθ (1 + ) θ = 0 dr rdr r The general solution of (1) is θ = 1 ν + θ θ θ (B1) () r C I () r C K () r ν From www. -efunda.-com/-math math Consider d θ dr + 1 r dr dθ m θ = 0 Let r =mr, it becomes d (B1 ) θ 1 d 0 + θ θ = dr r dr which is a modified Bessel equation of order zero, with the general solution of the form θ(r) = C 1 I 0 (mr) + C K 0 (mr) (B1a)
d θ 1 dθ m θ 1 θ θ 0 dr rdr Compare + =, having the general solution with θ(r) ) = C 1 I 0 (mr) + C K 0 (mr) d dx θ m θ = 0 which has the general solution θ ( x) = Ce + Ce mx 1 mx (B1a) (B) (Ba) Since the modified Bessel equation corresponds to the cylindrical coordinate, while () corresponds to the Cartesian coordinate, their solutions are similar in trend but (1a) reflects the effect of increasing surface area πr with increasing r.
3.6.5 Overall Surface Efficiency --characterizes an array of fins (Fig. 3.0) and the base surface to which they are attached q t = = q t η o (3.98) q max ha t θ b with A t = NA f + A b and q t = Nη f ha f θ b + ha b θ b, we obtain t f b t f f b b b η o = 1 NA f A t (1 η f ) (3.10)
If the fins are attached to the base, rather than an integral part of the wall, contact resistance should be included, as in Fig. 3.1 and Eqs. 3.104-105. 105. R toc, ( ) θ b = = q t η oc ( ) 1 ha (3.104) NAf η f η oc ( ) = 1 (1 ) A C (3.105) where C = 1 +η ha ( R / A ) t 1 f f t, c c, b 1 t η o = 1 NA f A t (1 η f ) EXs 3.10, 3.11